Let probability distribution is `[[x_i,:,1, 2, 3, 4, 5], [p_i,:,k^2, 2k, k, 2k, 5k^2]]` then value of `p(x gt 2)` is

Options

A. `1/6`
B. `7/12`
C. `1/36`
D. `23/36`

Answer

To solve the problem, we need to find the probability \( P(X > 2) \) given the probability distribution \( [[x_i,:,1, 2, 3, 4, 5], [p_i,:,k^2, 2k, k, 2k, 5k^2]] \).

### Step-by-Step Solution:

1. **Identify the Probability Distribution**:
The values of \( x_i \) are \( 1, 2, 3, 4, 5 \) and the corresponding probabilities \( p_i \) are \( k^2, 2k, k, 2k, 5k^2 \).

2. **Set Up the Equation for Total Probability**:
The sum of all probabilities must equal 1:
\[ k^2 + 2k + k + 2k + 5k^2 = 1 \]
Simplifying this gives:
\[ 6k^2 + 5k = 1 \]

3. **Rearrange the Equation**:
Rearranging the equation, we get:
\[ 6k^2 + 5k - 1 = 0 \]

4. **Solve the Quadratic Equation**:
We can use the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 6, b = 5, c = -1 \):
\[ k = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} \]
\[ k = \frac{-5 \pm \sqrt{25 + 24}}{12} \]
\[ k = \frac{-5 \pm \sqrt{49}}{12} \]
\[ k = \frac{-5 \pm 7}{12} \]
This gives us two possible values for \( k \):
\[ k = \frac{2}{12} = \frac{1}{6} \quad \text{and} \quad k = \frac{-12}{12} = -1 \]
Since probability cannot be negative, we accept \( k = \frac{1}{6} \).

5. **Calculate Probabilities for \( X > 2 \)**:
We need to find \( P(X > 2) \), which includes \( X = 3, 4, 5 \):
\[ P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) \]
Substituting the values:
\[ P(X = 3) = k = \frac{1}{6} \]
\[ P(X = 4) = 2k = 2 \cdot \frac{1}{6} = \frac{1}{3} \]
\[ P(X = 5) = 5k^2 = 5 \cdot \left(\frac{1}{6}\right)^2 = 5 \cdot \frac{1}{36} = \frac{5}{36} \]
Now summing these probabilities:
\[ P(X > 2) = \frac{1}{6} + \frac{1}{3} + \frac{5}{36} \]

6. **Find a Common Denominator**:
The common denominator for \( 6, 3, \) and \( 36 \) is \( 36 \):
\[ P(X > 2) = \frac{6}{36} + \frac{12}{36} + \frac{5}{36} = \frac{6 + 12 + 5}{36} = \frac{23}{36} \]

### Final Answer:
Thus, the value of \( P(X > 2) \) is:
\[ \boxed{\frac{23}{36}} \]